Wind Power Vehicle Traveling Down Wind Faster Than The Wind

[RCP]

@mender: I always like your basic explanations for how the cart works best. Find them reassuring in terms of how I perceive it. (And I wouldn't confess this elsewhere for fear of casting doubts on your understanding. )

I know it's in the blog somewhere, but bet you know the gear ratio of BB. What was it again? And using your explanation above, what would be the max BB could maintain steady state above ws? I may have seen this laid out many times before, but after 4k pages of such forgive an old guy if it overloads my RAM. The info is there, but often inaccessible.

After much thought I can finally envision BB leaving the packing popcorn in the dust. With spork on board we are talking about over 600 lbs.. Couldn't it store a lot of momentum under the right conditions getting up to ws? If a tin can will exceed ws for a few seconds in a lull after a gust, wouldn't BB be capable of doing so for ten seconds or more?

I'm speaking hypothetically above, as I know the NALSA rules take gusts and direction into consideration. But it was a first of its kind test for a sport's club, so might it not be in the realm of possibility they got some things wrong?

I would still love to see how I Ratant's cart would do on a treadmill after seeing that popcorn sail past it in natural wind. And remember, he was clever enough to design a cart that had remote steering and could ride the wind without needing a guide wire.

 

[chingel]

Thank you for your explanations.

Here is where I still have a problem. Doesn't energy matter also? Yes we can use gears to multiply the force, but doesn't the energy stay the same? Eg isn't the energy you put into at one end the same what you get from the other end, no matter what gears you use between? Otherwise you could just connect some gears to a wheel and then connect it to another wheel and multiply the force. You could do that, but doesn't the energy stay the same?

To me it seems that it takes a certain amount of energy to make the propeller move the air and that energy can only come from the kinetic energy of the cart, no matter what gears you use. If you use gears to make the propeller slower, the force at the wheels is lower, but since they move more in the same time, don't they expend the same amount of energy as the propeller gets?

 

[mheslep]

The air stream 'moves' the propeller more than the other way around. Or, in terms of energy, the propeller takes kinetic energy from the air stream and converts it to the kinetic energy of the platform.

 

[mender]

I know it's in the blog somewhere, but bet you know the gear ratio of BB. What was it again? And using your explanation above, what would be the max BB could maintain steady state above ws? I may have seen this laid out many times before, but after 4k pages of such forgive an old guy if it overloads my RAM. The info is there, but often inaccessible.

IIRC, the advance ratio is 0.8, and to get the theoretical top speed, you use this formula: 1/(1 - advance ratio). So for BB it would be 1/(1 - 0.8) or 1/0.2, which is five times the speed of the wind. Speaks pretty highly of the team when they managed to get runs in the 70% of TTS (theoretical top speed) range with their first vehicle on their second outing.

After much thought I can finally envision BB leaving the packing popcorn in the dust. With spork on board we are talking about over 600 lbs.. Couldn't it store a lot of momentum under the right conditions getting up to ws? If a tin can will exceed ws for a few seconds in a lull after a gust, wouldn't BB be capable of doing so for ten seconds or more?

So if a heavy tin can could beat the popcorn just by being at ws when the wind dies slightly, why do you have a hard time visualizing BB beating the (maybe) 10 mph popcorn when it is going almost 20 mph faster even without the wind dying? The only chance the popcorn (or a leaf) would have of beating a DDWFTTW vehicle is if a sudden gust of wind propelled the popcorn (or leaf) to speed before the cart has a chance to accelerate, making the BB's momentum a disadvantage.

I would still love to see how I Ratant's cart would do on a treadmill after seeing that popcorn sail past it in natural wind. And remember, he was clever enough to design a cart that had remote steering and could ride the wind without needing a guide wire.

As was the one in the original video by Goodman. Tested and proven as a TM cart, and would have beaten the popcorn handily even if they started at the same time.

It's nice (and still fun!) to run through this without the waters being muddied.

 

[mender]

Thank you for your explanations.

Here is where I still have a problem. Doesn't energy matter also? Yes we can use gears to multiply the force, but doesn't the energy stay the same? Eg isn't the energy you put into at one end the same what you get from the other end, no matter what gears you use between? Otherwise you could just connect some gears to a wheel and then connect it to another wheel and multiply the force. You could do that, but doesn't the energy stay the same?

To me it seems that it takes a certain amount of energy to make the propeller move the air and that energy can only come from the kinetic energy of the cart, no matter what gears you use. If you use gears to make the propeller slower, the force at the wheels is lower, but since they move more in the same time, don't they expend the same amount of energy as the propeller gets?

Yes, absolutely! That's what we've been talking about the whole time, the fact that the energy input from the wheels is used by the prop, and as such the energy output can never be more than the input. The force output certainly can, just like a lever and that's what we've also been talking about levers.

As mheslep pointed out, when the prop pushes air back relative to the cart it is reducing the speed of the air over the ground (reducing the kinetic energy of the wind) and using that energy to propel the cart (increasing the cart's kinetic energy). Since a real world cart is never 100% efficient, the cart will never reach the theoretical top speed but instead will travel at some percentage of it.

The Blackbird reached 3.5x the wind speed, which is 70% of its theoretical top speed of 5x wind speed; pretty impressive!

 

[rcgldr]

Doesn't energy matter also?

This was covered in the previous threads. Note that

power = force x speed

For the wheels, the point of application of force is the ground, which is moving backwards relative to the cart. For the propeller, the point of application of force is the air, which is moving at (ground speed relative to cart) - (wind speed relative to ground) = wind speed relative to cart.

As an example, say the wind speed relative to ground is +10 mph, and that the cart is moving at +25 mph (downwind). Ground speed relative to the cart is -25 mph. Wind speed relative to the cart is -15 mph. If the effective advance ratio is .8, then the propeller would produce thrust at -20 mph relative to the cart if there was no load. This ratio means that with zero losses, the force at the propeller can be 1.25 (25/20) times the opposing force at the wheels. Assume the force at the wheels is 80 lbs, then the propeller could produce up to 100 lbs of thrust with no losses:

power = 80 lbs x 25 mph (wheels) = 100 lbs x 20 mph (propeller)

This results in an ideal net forward force of 20 lbs. The real thrust force and speed will be less, but as long as the net force is greater than rolling resistance and drag the cart accelerates, and in the case of the BB cart, it reaches a max around 3.5x wind speed.

The propeller makes things a bit more complicated than just gearing. The propeller's pitch is part of the effective gearing (and advance ratio). The propeller's size (width and length) affects how much thrust force is produced for a given thrust speed (relative to the air) and pitch. A long (large diameter) propeller will generally be more efficient.

 

[chingel]

I'm just a regular student and to be honest I never quite understood how power = force x speed. I have always thought that power means how many joules in a second you can do. The speed of what it is in the formula? When it takes me certain energy to produce a force of 1 N and I apply the force on a 0,5 kg object it moves faster than when I apply the same force on a 1 kg object. It's not like I double my power output, isn't it? Because I am using the same energy per second, the object just offers less resistance and moves faster, in no way it shouldn't increase my power output in my understanding.

I guess I am stuck on the energy concept. Is there a way to explain it in terms of energy? The propeller and the wheels are connected, energy input has to equal energy output, the energy that it takes from the kinetic energy of the cart it gives to the air particles trying to propel itself, but since the energies are equal they should cancel out, energy shouldn't change no matter what gears you use. Some extra energy has to come from somewhere.

 

[A.T.]

I'm just a regular student and to be honest I never quite understood how power = force x speed.

It's just the time derivate of:

work = force x distance

since:

power = work / time
velocity = distance / time

you have:

power = force x velocity

This form allows you to analyze energy balance instantaneously, which is handy for analyzing accelerating things.

I guess I am stuck on the energy concept. Is there a way to explain it in terms of energy?

First you have to pick an inertial reference frame to do your energy analysis using work = force x distance. For the energy balance in the reference frame of the cart check this video at 1:05:

energy input has to equal energy output

But force output can greater that force input.

The propeller and the wheels are connected, energy input has to equal energy output, the energy that it takes from the kinetic energy of the cart it gives to the air particles trying to propel itself,

About which reference frame are you talking here?

 

[A.T.]

The propeller makes things a bit more complicated than just gearing. The propeller's pitch is part of the effective gearing (and advance ratio).

Here is a diagram that shows how propeller pitch and advance ratio are connected:

 

[poont2]

OMG this diagram just blew up my brain, lol. what is that gear and what is that grey vertical teeth? Let me spend some time to understand it, I guess I have less than half of a brain of normal human...

But one thing I am wondering, when the car with propeller is stationary with no wind, and wind start to blow (same direction and the car's direction). If the propeller is not connected to the wheel, wouldn't the propeller spin in the opposite angular direction that you want it to spin? does this meen that the force from the wheel would greatly overpower the this effect? I am sure if that make sense...

 

[rcgldr]

What is that gear and what is that grey vertical teeth?

The gear represents the wheels. The grey vertical bar represents the drivetrain connecting the gear (wheels) and the propeller.

When the car with propeller is stationary with no wind, and wind start to blow ...

For a downwind cart in that situation the propeller act's as a "bluff body" (a sail), blocking the wind. This results in a forward force on the cart, which results in an backwards force (torque) on the wheels, and the turning wheels torque is passed through the drivetrain to cause the propeller to turn.

One exception case is if there isn't enough grip at the wheels to handle the wind or a sudden gust, where it's possible for the cart to move forwards, but with the propeller causing the wheels to spin (slide) backwards. As the cart picks up speed, this reverse torque from the propeller will diminish, and eventually the wheels will regain grip unless the grip factor is very low.

 

[A.T.]

OMG this diagram just blew up my brain, lol. what is that gear and what is that grey vertical teeth?

As rcgldr said, the gear is the wheel, the gray rack is the transmission, that couples wheel and propeller rotation. It is important to point out, that the vertical direction in the diagram, corresponds to the circumferential direction for the airfoil, which is spinning not just going up.

But one thing I am wondering, when the car with propeller is stationary with no wind, and wind start to blow (same direction and the car's direction). If the propeller is not connected to the wheel, wouldn't the propeller spin in the opposite angular direction that you want it to spin?

Correct. See the wheel rotation for CASE C - HELD IN AIR: the wheels have no ground contact so the rotor is freewheeling as a turbine and turns the wheels "backwards" (as if the cart would go upwind)

does this meen that the force from the wheel would greatly overpower the this effect? I am sure if that make sense...

Yes, the torque transmitted from the wheels is greater than the aerodynamic torque, so the rotor turns as a propeller, against the aerodynamic torque.

You can also put it the other way around, like it is shown in the diagram (CASE C - START UP): The axial aerodynamic force on the propeller(blue arrow) that pushes the cart downwind is greater than the ground reaction force (red arrow) caused by the circumferential aerodynamic force(or torque) on the propeller.

In reality the axial aerodynamic force on the propeller(blue arrow) is of course also supported by the aerodynamic force (bluff body drag) on the chassis, supporting the self start.

 

[chingel]

Regarding power = force x speed. Is the speed in the equation the speed of the object that force is applied to? Doesn't the speed of the object depend on how much resistance it offers to the force, not on the power? If I have a spaceship that uses a laser on Earth for propulsion using a light sail, the laser doesn't know or care how fast the ship is going, to him it doesn't even matter if it hits anything, it is just outputting at a constant power, the ship speeds up, but the power doesn't change. Basically I just don't understand the equation.

The force at the propeller is bigger, but since it is moving slower, it uses up as much energy as the wheels give. You can't just use gears to do more work, where does the cart get the extra energy? This thing is frying my brain.

When you are going downwind and reach wind speed the tangential airflow component and the downwind force component is zero, so they shouldn't increase the speed. Instead the wheels start driving the propeller, slowing the cart down. How does it work?

 

[A.T.]

Regarding power = force x speed. Is the speed in the equation the speed of the object that force is applied to?

Yes

Doesn't the speed of the object depend on how much resistance it offers to the force,

No. The instantaneous speed doesn't. The instantaneous acceleration does.

If I have a spaceship that uses a laser on Earth for propulsion using a light sail, the laser doesn't know or care how fast the ship is going, to him it doesn't even matter if it hits anything, it is just outputting at a constant power, the ship speeds up, but the power doesn't change.

The power in power = force x speed here is the change of the ships kinetic energy per time, not the laser power. Since KE grows with velocity squared, the input of KE increases even if the acceleration is constant.

The force at the propeller is bigger, but since it is moving slower, it uses up as much energy as the wheels give.

In the cart frame, the air moves slower than the ground. So the cart can do negative work on the fast ground (harvest energy) with a low force, and use that energy to do positive work on the slow air with a large force. The force difference is the net thrust.

In the ground frame the air is doing positive work on the cart, while the cart uses the ground as a fulcrum to multiply the airs speed.

https://www.youtube.com/watch?v=g8bxXRQtcMY

You can't just use gears to do more work,

Correct. But you can use gears to get more speed:

https://www.youtube.com/watch?v=E7vcQcIaWSQ

where does the cart get the extra energy?

Which extra energy? You have not even stated in which reference frame you are doing your energy analysis?

When you are going downwind and reach wind speed the tangential airflow component and the downwind force component is zero,

Wrong. At wind speed the propeller is spinning in still air, so the airflow at the blade is mostly tangential. The airfoil generates lift perpendicular to the airflow so the downwind force component is not zero.

Instead the wheels start driving the propeller, slowing the cart down.

The wheels are turning the propeller right from the start, not from windspeed on. But they are not slowing the cart down, because the prop's axial force is greater than their retarding force.

 

[chingel]

The power in power = force x speed here is the change of the ships kinetic energy per time, not the laser power. Since KE grows with velocity squared, the input of KE increases even if the acceleration is constant.

Does this mean that you cannot calculate the power of the laser knowing the force the laser applies and the speed at which the ship moves? Does it also mean that the rate of growth of the ships kinetic energy will be exponential as the speed keeps increasing? Why doesn't the equation using the propeller's thrust and windspeed apply to the change of the air particles kinetic energy per time instead of the propellers power? The equation is very confusing to me.

In the video the cart certainly does move faster than the blue chain if it has a rod stuck in the moving chains, but I cannot understand how it is the same with the DDWFTTW.

If the cart is at windspeed, the wind doesn't push the propeller or the cart, the propeller gets moved only by the wheels. This means that any energy that the propeller uses when turning comes out of the wheels and thus the carts kinetic energy. When you are driving a car you can't just use the moving ground to do negative work and use gears to lever it, any energy you extract will slow you down exactly as much. Electric cars use breaking energy to generate electricity, they can't just do it while cruising without energy loss. There has to be some other effect at play with the wind cart because they somehow can go faster than the wind and I cannot understand what it is.

 

[A.T.]

Does this mean that you cannot calculate the power of the laser knowing the force the laser applies and the speed at which the ship moves?

I think you would at least have to know how much of the light is reflected. But you cannot assume that the increase of the ships KE per time equals the laser power.

Does it also mean that the rate of growth of the ships kinetic energy will be exponential as the speed keeps increasing?

Classicaly KE grows with speed squared, not exponentially.

Why doesn't the equation using the propeller's thrust and windspeed apply to the change of the air particles kinetic energy per time instead of the propellers power?

In the ground frame the equation does apply as you suggest:

power_air_to_cart = true_wind_speed x thrust

In the cart's frame:

power_ground_to_cart = ground_speed x wheel_drag
power_cart_to_air = air_speed x thrust

The equation is very confusing to me.

You have to realize that KE is a frame dependent quantity. Every explanation based on KE will be valid only for a certain frame. A different observer will see a completely different energy balance. So maybe it is not that useful to ask for "explanations in terms of energy". But if you do, you should at least specify the reference frame.

In the video the cart certainly does move faster than the blue chain if it has a rod stuck in the moving chains, but I cannot understand how it is the same with the DDWFTTW.

Put a big sail in the middle of the rod, stick one end into the ground and the other end will go twice the windspeed.

If the cart is at windspeed, the wind doesn't push the propeller

Why not? The blades are spinning. They have a relative airflow perpendicular to true wind direction. It is trivial to produce a downwind force with an airfoil set at the right angle. And with an efficient airfoil that downwind force can be 20 times greater than the tangential drag of the blade, that brakes the wheels.

the propeller gets moved only by the wheels.

That's is true in the cart's frame, where the propeller is only rotating. Here the ground moves, and turns the propeller via the wheels.

In the ground frame you can see the wheels & transmission as mere kinematic constraint for the blades, that enforce a certain helical path for the airfoils. The blades are pushed by the air along that helical path:

[PLAIN]http://img811.imageshack.us/img811/4922/propellervectors.png

Here for comparison a sailcraft on broad reach, constrained laterally by the keel/skates/wheels:

[PLAIN]http://img253.imageshack.us/img253/6694/downwindvectorsen3.png

Here a nice animation that transforms one into the other:

https://www.youtube.com/watch?v=UGRFb8yNtBo

This means that any energy that the propeller uses when turning comes out of the wheels and thus the carts kinetic energy.

Your statements about energy are meaningless without specifying which reference frame you mean.

In the cart's frame the cart doesn't have any KE. But the ground has plenty of it and gives it to the air, via wheels and propeller.

In the ground frame the propeller is taking KE from the air, so it is an energy input to the cart. To confirm this check the vectors of blade_velocity and blade_force in the diagram above. They are at less than 90° (their dot product is positive), so the air is doing positive work on the blade. Check also the red tracer in the very first clip of the below animation. It shows how the air gets slowed down in the ground frame.

https://www.youtube.com/watch?v=FqJOVHHf6mQ

When you are driving a car you can't just use the moving ground to do negative work and use gears to lever it, any energy you extract will slow you down exactly as much.

You will not slow down, if you use the extracted energy to push against something that moves slower relative to you than the ground. The net force determines if you slow down, not just the force from the ground.

There has to be some other effect at play with the wind cart because they somehow can go faster than the wind and I cannot understand what it is.

There is no "other effect". It is just gearing. You should try to understand rigid examples of such gearing first:

https://www.youtube.com/watch?v=k-trDF8Yldc

 

[rcgldr]

If I have a spaceship that uses a laser on Earth for propulsion using a light sail, the laser doesn't know or care how fast the ship is going ...

From the Earth's frame of reference, as the speed of the spaceship increases, the number of photons per second that hit the spaceship "sail" decreases.

If the cart is at windspeed, the wind doesn't push the propeller or the cart, the propeller gets moved only by the wheels.

The wind pushes against the air being propelled by the propeller. That air just aft of the propeller experiences some amount of compression, generating equal and opposing forces, upwind againts the true wind, and downwind against the propeller. That downwind force on the propeller corresponds to the thrust generated by the propeller, which due to the effective gearing (advance ratio) produces this thrust at a lower speed (relative to the cart), than the force at the wheels from the ground. This allows the thrust to be greater than the opposing force from the ground while at the same time using less power because of the reduced speed of that thrust. The reduced speed is possible because the wind speed is always less than the ground speed when the cart is moving downwind, regardless of the speed of the cart (relative to the ground).

 

[Redbelly98]

It's meant to be an exercise in aerodynamics. Break the problem down into a free body diagram and figure out if it is possible or not.

Better yet, let the car be moving at the same speed as the wind, so that the air drag is zero. Then figure out if the acceleration can be positive, in which case faster than the wind is possible.

 

[OmCheeto]

Better yet, let the car be moving at the same speed as the wind, so that the air drag is zero. Then figure out if the acceleration can be positive, in which case faster than the wind is possible.

Oh! I never thought of that. Quite a while back, I ran a computer simulation on the device from a standstill, and it only reached 70% of wind speed. Though it was a fixed prop pitch simulation. It appears that the full scale model has variable pitch blades. Back to the drawing board.

 

[ThinAirDesign]

With spork on board we are talking about over 600 lbs.. Couldn't it store a lot of momentum under the right conditions getting up to ws? If a tin can will exceed ws for a few seconds in a lull after a gust, wouldn't BB be capable of doing so for ten seconds or more?

RCP, if you'll try to remember (or even read the NALSA rules again), you will see that during the 10 second timing period the BB cart had to be *accelerating* -- that is it had to exit the timing period *faster* than it entered. The tin can that is beating the wind during a quick lull is doing *just the opposite* -- it is slowing down.

The requirement for acceleration through the timing period was added *precisely* to make it impossible for the BB to 'coast' through a lull to get the record.

so might it not be in the realm of possibility they got some things wrong?

Like what?

I would still love to see how I Ratant's cart would do on a treadmill after seeing that popcorn sail past it in natural wind. And remember, he was clever enough to design a cart that had remote steering and could ride the wind without needing a guide wire.

OMG RCP -- "clever enough"? You think the reason we didn't put RC on our small model was because we weren't "clever" enough?. That's freakin' funny. ROFLAO. Goodman put RC on and was accused of using the battery that was onboard to power the cart. THAT'S why we never made one RC. We wanted to be able to say correctly that there were no batteries on board at all.

But then you've known this over and over -- you just can't keep it all together.

JB

 

[A.T.]

Oh! I never thought of that. Quite a while back, I ran a computer simulation on the device from a standstill, and it only reached 70% of wind speed. Though it was a fixed prop pitch simulation. It appears that the full scale model has variable pitch blades. Back to the drawing board.

Their first prototype had fixed pitch and went above 2x windspeed from a standstill. but the acceleration below windspeed was very low. Maybe your pitch was too high, the wind too low or the assumed inefficiencies worse then in reality. The self start is not really the key claim here anyway. The core question is if it can achieve a steady state above windspeed.

How did you calculate the propeller axial force and reaction torque below windspeed? I did some simulations using http://www.mh-aerotools.de/airfoils/javaprop.htm" libraries (propeller simulation based on blade element theory). This software cannot really handle the the situation at windspeed(zero airspeed) very accurately so I started the cart just above windspeed. I used the actual propeller geometry of the Blackbird and with the folowing parameters:

transmission eff: 0.93
aero drag coefficient: 0.22
rolling drag coefficient: 0.01
frontal area[m^2]: 2
mass[kg]: 295

it went close to 3 x windspeed in 4.5m/s wind.

WS = true windspeed
wheel_drag = retarding force on wheel needed to turn the prop
net_thrust = prop_thrust - wheel_drag
net_friction = aero_drag + rolling_friction

The pitch was variable and adjusted for best acceleration (see blade_angle bottom left)

 

[A.T.]

transmission eff: 0.93
aero drag coefficient: 0.22
rolling drag coefficient: 0.01
frontal area[m^2]: 2
mass[kg]: 295

it went close to 3 x windspeed in 4.5m/s wind.

Since the efficiency parameters are estimates I also checked the sensitivity of the result to those parameters:

 

[OmCheeto]

Maybe your pitch was too high, the wind too low or the assumed inefficiencies worse then in reality.

Way too high according to your chart. I had a pitch of 45 degrees, based probably on what I could visually observe from the original mini-carts.

How did you calculate the propeller axial force and reaction torque below windspeed?

Most embarrassing. I searched for over 2 hours yesterday and could not find the simulation. But anyways, it's moot point, now that we know my prop pitch was incorrect.

 

[A.T.]

Way too high according to your chart. I had a pitch of 45 degrees, based probably on what I could visually observe from the original mini-carts.

The plotted angle is the blade angle at 75% prop radius. Since the blade is twisted it is much higher closer to the root. The transmission ratio between propeller_omega and ground_speed plays also a role here.

 

[Fun Value]

Well, I guess I don't have half-decent brain power because I thought about for a few minutes and couldn't come up with it. Damn my less than half decent brain...

Same with me. I guess we should have more ego.
On the other hand, I think the original post suggested downwind, I think implying no angle, so it did seem interesting. I am to busy to spend much time thinking about it, and if someone can clearly and simply explain without trying to sound like a genius, I would like to hear about it.

 

[mender]

Essentially, it's an overdrive gearing system that uses two inputs (air speed and ground speed) to generate an output (cart speed).

See A.T.'s last post on page three and start with the "Under the Ruler Faster Than the Ruler" video.

 

[chingel]

I think you would at least have to know how much of the light is reflected. But you cannot assume that the increase of the ships KE per time equals the laser power.

Just for the sake of the argument let's assume 100% efficiency.

In the cart's frame:

power_ground_to_cart = ground_speed x wheel_drag
power_cart_to_air = air_speed x thrust

Are these powers equal as I understand from the previous posts? This means that the propeller takes as many joules from the cart's kinetic energy as it gives to the air particles.

That's is true in the cart's frame, where the propeller is only rotating. Here the ground moves, and turns the propeller via the wheels.

But you can't just have the ground turn the propeller without slowing the car down by an equal amount. You can't use the moving ground when driving a car to turn a generator without slowing the car down.

The diagrams I don't understand either. They already have 1,5 windspeed at the beginning? What happens at windspeed? The air is still, the propellers get turned by the wheels, which slows down the cart, the propeller speeds up the cart. With 100% efficiency the speed should stay the same.

From the Earth's frame of reference, as the speed of the spaceship increases, the number of photons per second that hit the spaceship "sail" decreases.

Yes, but what does it mean? The ship still keeps speeding up and the laser's power stays the same, doesn't it? Let's assume 100% efficiency at using the photons for propulsion. If the ship goes from 1 km/h to 2 km/h, the power of something doubles due to the formula, but the power of the input engine ie the laser doesn't change. The effect of the small speed increase on the amount of photons hitting is negligible How does the formula exactly work?

That downwind force on the propeller corresponds to the thrust generated by the propeller, which due to the effective gearing (advance ratio) produces this thrust at a lower speed (relative to the cart), than the force at the wheels from the ground. This allows the thrust to be greater than the opposing force from the ground while at the same time using less power because of the reduced speed of that thrust. The reduced speed is possible because the wind speed is always less than the ground speed when the cart is moving downwind, regardless of the speed of the cart (relative to the ground).

When you put gears in there to make the propeller move slower, it displaces less air and also drags the wheels less. I can't see how it can bypass taking energy from the cart to propel the air particles. They are standing still to the propeller and to move them it needs energy from the cart, no matter what gearing you use to make it slower or faster.

 

[rcgldr]

The wind pushes against the air being propelled by the propeller. That air just aft of the propeller experiences some amount of compression, generating equal and opposing forces, upwind againts the true wind, and downwind against the propeller. That downwind force on the propeller corresponds to the thrust generated by the propeller, which due to the effective gearing (advance ratio) produces this thrust at a lower speed (relative to the cart), than the force at the wheels from the ground. This allows the thrust to be greater than the opposing force from the ground while at the same time using less power because of the reduced speed of that thrust. The reduced speed is possible because the wind speed is always less than the ground speed when the cart is moving downwind, regardless of the speed of the cart (relative to the ground).

When you put gears in there to make the propeller move slower, it displaces less air and also drags the wheels less. I can't see how it can bypass taking energy from the cart to propel the air particles. They are standing still to the propeller and to move them it needs energy from the cart, no matter what gearing you use to make it slower or faster.

The input power = (ground speed relative to cart) x (ground force on the wheels)
The output puter = (thrust speed relative to cart) x (thrust force from the propeller)

Since ground speed > thrust speed, then ground force can be < thrust force, even after losses due to power conversion. I already gave an example with some made up number in post #36:

post 36

 

[chingel]

And they are equal as I understand from the post 36? The joules per second it gets from the wheels is the joules per second the propeller uses to propel the air? Why doesn't this mean that any energy that the propeller gives to the air particles trying to propel itself, no matter how fast they are, no matter what gears you use, gets taken from the carts kinetic energy and therefore slows the cart down?

 

[mender]

Because the energy to power the cart is from the wind. After the cart passes, the kinetic energy of the wind is lower; the prop slows down the speed of the air mass (wind) relative to the ground.

If there wasn't a wind, you'd be right. The wind is the external energy source; no wind = no energy source to power the cart.